Question

An aqueous solution has 5% urea and 10% glucose by weight. What will be the freezing point of this solution? (K$_{f}$ water = 1.86 K kg mol$^{-1}$)

Answer 1

270.11 K

Answer 2

To find the freezing point of the solution, we first need to calculate the molality of the solution. The solution contains 5% urea and 10% glucose by weight. This means that in 100 g of the solution, there are 5 g of urea and 10 g of glucose. The remaining weight is the solvent, water.

Weight of solvent (water) = 100 g - (Weight of urea + Weight of glucose)
Weight of solvent = 100 g - (5 g + 10 g) = 100 g - 15 g = 85 g

The molecular weight of urea ($CO(NH_2)_2$) is approximately 60 g/mol.
The number of moles of urea ($n_{urea}$) = $\frac{\text{Weight of urea}}{\text{Molecular weight of urea}} = \frac{5 \text{ g}}{60 \text{ g/mol}} = \frac{1}{12}$ mol.

The molecular weight of glucose ($C_6H_{12}O_6$) is approximately 180 g/mol.
The number of moles of glucose ($n_{glucose}$) = $\frac{\text{Weight of glucose}}{\text{Molecular weight of glucose}} = \frac{10 \text{ g}}{180 \text{ g/mol}} = \frac{1}{18}$ mol.

Since both urea and glucose are non-electrolytes, the total number of solute particles depends on the total number of moles of urea and glucose.
Total number of moles of solute ($n_{total}$) = $n_{urea} + n_{glucose} = \frac{1}{12} + \frac{1}{18}$ mol.
To add these fractions, we find a common denominator, which is 36.
$n_{total} = \frac{1 \times 3}{12 \times 3} + \frac{1 \times 2}{18 \times 2} = \frac{3}{36} + \frac{2}{36} = \frac{3+2}{36} = \frac{5}{36}$ mol.

Molality ($m$) is defined as the number of moles of solute per kilogram of solvent.
Weight of solvent in kg = $\frac{85 \text{ g}}{1000 \text{ g/kg}} = 0.085$ kg.
Molality ($m$) = $\frac{n_{total}}{\text{Weight of solvent in kg}} = \frac{5/36 \text{ mol}}{0.085 \text{ kg}} = \frac{5}{36 \times 0.085} = \frac{5}{3.06}$ mol/kg.
$m = \frac{500}{306} = \frac{250}{153}$ mol/kg.

The depression in freezing point ($\Delta T_f$) is given by the formula:
$\Delta T_f = K_f \times m$
Given $K_f$ for water = 1.86 K kg mol$^{-1}$.
$\Delta T_f = 1.86 \times \frac{250}{153}$ K.
$\Delta T_f = \frac{1.86 \times 250}{153} = \frac{465}{153}$ K.
$\Delta T_f = \frac{155 \times 3}{51 \times 3} = \frac{155}{51}$ K.
$\Delta T_f \approx 3.0392$ K.

The normal freezing point of water is 0$^\circ$C, which is equal to 273.15 K.
The freezing point of the solution ($T_f$) is given by:
$T_f = \text{Normal freezing point of water} - \Delta T_f$
$T_f = 273.15 \text{ K} - \frac{155}{51} \text{ K}$
$T_f = 273.15 - 3.039215686 \approx 270.11078$ K.

If the normal freezing point of water is taken as 273 K (as sometimes used in simplified calculations), then:
$T_f = 273 \text{ K} - 3.0392 \text{ K} \approx 269.9608$ K.

Using the more precise value for the freezing point of water (273.15 K) and the calculated $\Delta T_f$, the freezing point of the solution is approximately 270.11 K. This matches the value obtained in the similar question's final result, suggesting that 273.15 K was implicitly used as the normal freezing point of water.

The freezing point of the solution is approximately 270.11 K.
If the answer is required in $^\circ$C, it would be $0 - \Delta T_f = -3.0392^\circ$C.